evaluate the sum of the finite geometric series. 1+2+4+8+...+128
S = a*((1 - r^(n+1))/(1-r)) S = 1*((1 - 2^(7+1))/(1-2)) S = 1*((1 - 2^8)/(1-2)) S = 1*((1 - 256)/(1-2)) S = 1*((-255)/(-1)) S = 255 So 1+2+4+8+...+128 = 255
what is the original equation? So i can use it on the other problems
If you have the geometric sequence, a[n] = a*r^(n-1), then the sum from the first term 'a' to the nth term a*r^(n-1) is S = a*((1 - r^n)/(1-r))
Ops. i read infinite. lol
@luisrock2008 , http://2.bp.blogspot.com/_fTmELBvhTaM/TEbU_1iaqYI/AAAAAAAABQo/d6d2J1dyPuQ/s1600/formula.png
for powers of two you can double the last number and subtract one \[1+2+2^2+...+2^n=2^{n+1}-1\] so in your case you can take 128, double it and get 256, subtract 1 get 255
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